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STRUCTURE OF P-32 AND S-32
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper ““Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS. Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Structure of the unstable P-32 with S = +1. Why P-32 turns to the stable S-32 ''' In the first diagram one can see the core of P-32 (from p1 to n14) which represents the structure of Si-28 with S=0. (See my papers of the nuclear structure in my Fundamental Physics Concepts). The core consists of the elongated parallelepiped with six horizontal squares of S =0 of Mg-24 and the two rectangles with S=0 formed by the nucleons p13, n13 , p14 and n14. Fortunately these two rectangles are able to form the two blank positions in which the additional neutrons as n15(+1/2) and n16(+1/2) with S = +1 can form the stable four np bonds respectively, because at each point of n15 and n16 we observe two np bonds per neutron. For example at n15 we observe the two bonds as n15p9 and n15p13. It is of interest to note that P-32 consists of 15p and 17n. Fortunately a favorable blank position for the formation of the two pn bonds of the additional p15 can exist at the point p15, where we observe two pn bonds as the p15n3 and p15n13. Also n17 is able to form the two np bonds per nucleon as the axial bond n17p15 and the radial bond n17 p1. However the radial np bond has a weak binding energy because at p1 there are four pn bonds per nucleon. For example the n13p5 bond is stronger than the n17p1 bond because at p5 there exist five bonds per nucleon which contribute to the increase of the binding energies. Under this condition the n17 turns to a proton which represents the p16 of the stable S-32. Since the spin of the Mg-24 =0 one concludes that the total spin S= 0 is due to the spin S= 0 of all additional nucleons from p13 to n16. It is of interest to note that the two rectangles of the structure of Si-28 contribute to the formation of two blank positions for the n15 and n16. The two rectangles also contributr to the formation of two blank positions for making a stable structure with p15 and p16. '''Unstable P-32 with S = +1 Stable S-32 with S = 0 ' ' ' p12(-1/2)n12(-1/2) p12(-1/2)n12(-1/2)' ' n11(-1/2)p11(-1/2) n11(-1/2)p11(-1/2) ' ' n10(+1/2)p10(+1/2)…..n16(+1/2) n10(+1/2)p10(+1/2)....n16(+1/2)' ' n15(+1/2 …. p9(+1/2)n9(+1/2) n15(+1/2)… p9(+1/2)n9(+1/2)' ' p8( -1/2)n8( -1/2) ….p14(-1/2) p8( -1/2) n8 (-1/2)….p14(-1/2) ' ' p13(-1/2) …n7( -1/2)p7(-1/2) p13(-1/2) …n7( -1/2)p7( -1/2) ' ' n6(+1/2)p6(+1/2)… n14(+1/2) n6(+1/2)p6(+1/2)….n14(+1/2)' ' n13(+1/2)….p5(+1/2)n5(+1/2) n13(+1/2)…p5(+1/2)n5(+1/2) ' ' p4( -1/2) n4(-1/2) p4(-1/2) n4(-1/2)…...p16(-1/2)' ' p15(-1/2) …n3(-1/2)p3(-1/2) p15(-1/2)… n3(-1/2)p3(-1/2)' ' n2(+1/2)p2(+1/2) n2(+1/2)p2(+1/2) ' ' n17(+1/2) …..p1(+1/2)n1(+1/2) p1(+1/2)n1(+1/2)' ' ' 'Stable S-32 with S = 0. ' In the second diagram one sees that S-32 has the same core as that of P-32. That is, it consists of Mg-24 with S=0 and the two rectangles with S = 0 which form the two blank positions for the two neutrons n15(+1/2) and n16(+1/2). Thus the two extra neutrons n15 and n16 do not turn into protons because they have two bonds per nucleon. For example at n16 we observe the two np bonds as n16p10 and n16p14 . Fortunately for the additional p15(-1/2) the neutrons n4 and n14 form a new blank position able to receive the additional p15 which forms the two stable pn bonds per nucleon like p15n4 and p15n14. In other words the repulsive energy Q of long range cannot overcome the pn bonds and the P-31 has always a stable structure of S = +1/2. Since the core (Si-28) gives S= 0 one concludes that the S = +1/2 of P-31 is due to the additional nucleons. That is 'S = n15(+1/2) + n16(+1/2) + p15(-1/2) = +1/2. ' Category:Fundamental physics concepts